a/ $AD$ là đường phân giác $\widehat{BAC}$
$→\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{9}{6}=\dfrac{3}{2}$
$→3DC=2DB$
$→DC=\dfrac{2DB}{3}$
$DB+DC=BC$
$→\dfrac{2DB}{3}+DB=10$
$→\dfrac{5DB}{3}=10$
$→5DB=30$
$→DB=6$ (cm)
$→DC=4$ (cm)
b/ Kẻ đường cao $AH$ vuông góc $BC$
$S_{ΔABD}=\dfrac{1}{2}.AH.BD$
$S_{ΔACD}=\dfrac{1}{2}.AH.CD$
$→\dfrac{S_{ΔABD}}{S_{ΔACD}}=\dfrac{\dfrac{1}{2}.AH.BD}{\dfrac{1}{2}.AH.CD}=\dfrac{BD}{CD}=\dfrac{3}{2}$